pole

**Kat-M** · April 19th 2010, 01:18 AM

Let $R(z)$ be a rational function: $R(z)=p(z)/q(z)$ where $p$ and $q$ are holomorphic polynomials. Let $f$ be holomorphic on $C$ \ $\{P_1, P_2,....,P_k \}$ and suppose $f$ has a pole at each of the points $P_1, P_2,....,P_k$ . Finally assume that $|f(z)| \leq |R(z)|$ for all $z$ at which these functions are defined. Prove that $f$ is a constant multiple of $R$ .

**tonio** · April 19th 2010, 02:28 AM

Originally Posted by Kat-M

Let $R(z)$ be a rational function: $R(z)=p(z)/q(z)$ where $p$ and $q$ are holomorphic polynomials. Let $f$ be holomorphic on $C$ \ $\{P_1, P_2,....,P_k \}$ and suppose $f$ has a pole at each of the points $P_1, P_2,....,P_k$ . Finally assume that $|f(z)| \leq |R(z)|$ for all $z$ at which these functions are defined. Prove that $f$ is a constant multiple of $R$ .

You meant, I presume, that $\{P_1,\ldots,P_k\}$ are the zeros of the denominator polynomial $q(z)$ , right? Well, we get that $q(z)f(z)$ is an entire function and

$|q(z)f(z)|\leq |p(z)|\Longrightarrow$ as no entire function can dominate another entire function unless one is a scalar multiple of the other (this follows at once from

Liouville's Theorem), we get what we want.

Tonio

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